1 MA40S PRECALCULUS UNIT F CONICS CLASS NOTES (Completed) What Are Conic Sections? Objectives: To visualize the shapes generated from the intersection of a cone and a plane For each conic section, to describe the relationship between the plane, the central axis of the cone, and the cone s generator 1) The Cone Consider a right triangle with hypotenuse c, and legs a, and b. We can generate a 3 dimensional solid called a cone by rotating the triangle about its leg b. The leg a, generates a circle, which becomes the base of the cone. The leg c is called the generator of the cone. The line l is the axis of symmetry of the cone. The point A is called the vertex of the cone. Document1 Revised:
If we place together the vertices of two cones such that they share the same axis of symmetry, we have what is called a double-napped cone. Label the diagram with the following. 1) vertex A ) axis of symmetry 3) generator
3. Conic Sections When a plane intersects a cone, it can form two dimensional objects called conic sections. The type of conic section formed depends upon the angle at which the plane intersects the cone. Conic: Circle The plane is parallel to the base. Conic: Ellipse The plane is inclined towards the base of the cone. Conic: Parabola The plane is parallel to the generator of the cone. Conic: Hyperbola The plane is parallel to the axis of symmetry, and intersects both nappes of the cone. A hyperbola has two distinct branches.
4 The Circle Objectives: To investigate the standard form and the general form of the equation of a circle To determine the radius and centre of a circle given its equation To write the equation of a circle given its centre and radius 1. Review Recall from Principles of Math 11, and from Lesson 1, Chapter 3, the following definition of a circle. A circle is the set (or locus) of points in a plane that are equidistant from a fixed point. The fixed point is called the centre. The distance from the centre to any point on the circle is called the radius. Example 1: Write the equation of the circle with radius 5 and centre (3, ). Let P(x,y) be any point on the circle. Then, using the distance formula, we have (x 3) + (y ) = 5. The equation of the circle is (x 3) + (y ) = 5.. The Standard Form of the Equation of a Circle The following is the standard form of the equation of a circle with radius r, and centre (h,k). (x - h) + (y - k) = r
5 Example : Write the equation of the circle with radius 7 and centre ( 4,8). We have h = 4, k = 8, and r = 7. (x 4) + (y 8) = 7 (x + 4) + (y 8) = 49 3. The General Form of the Equation of a Circle If we expand the equation in Example, we obtain the following result. (x + 4) + (y 8) = 49 (x + 8x + 16) + (y 16y + 64) = 49 x + y + 8x 16y + 31= 0 For the circle in Example, the equation is said to be in the general form. The following is the general form of the equation of a circle. Ax + Ay + Dx + Ey + F = 0 where A, D, E, and F are real constants and A 0. Thus for the circle in Example, we have A = 1, D = 8, E = 16, and F = 31. Note: We are reserving the real constants B and C for the general forms of other conics. Example 3: Determine the radius and centre of the circle whose general form equation is x +10x + y 6y +18 = 0. To transform x +10x + y 6y +18 = 0 into standard form, which tells us the centre and radius of the circle, we use the process called completing the square. Centre: ( 5, 3) Radius: 4
6 x + 10x + y 6y + 18 = 0 x + 10x + y 6y = 18 (x + 10x + 5) + (y 6y + 9) = 18 + 5 + 9 (x + 5) + (y 3) = 16 Example 4: If m and n are positive integers, determine the radius of the circle mx + my n = 0. mx + my n = 0 x + y = n m Thus r = n m Example 5: Graph the equation (x + 5) + ( y 3) = 16 using a graphing calculator. (Use the Square viewing window Zoom Square.) Sketch the result on the grid below. First, solve the equation for y in terms of x. (x + 5) + (y 3) = 16 (y 3) = 16 (x + 5) y 3 = ± 16 (x + 5) y = ± 16 (x + 5) + 3 This means we must graph two functions: Y 1 = + 16 (x + 5) + 3 and Y = 16 (x + 5) + 3
7 The Ellipse Objectives: To investigate the standard form and the general form of the equation of an ellipse To graph an ellipse given its equation To write the equation of an ellipse given its graph 1. Stretching the Graph of a Circle Exercise 1: Let s compare the graph of the circle x + y = 1 to the graph of Recall the rules for stretching a function y = f (x). x 3 + y =1. The function y a = f (x) [or y = af (x) ] can be obtained by stretching the function y = f (x) according to the following rules. If a > 1, there is a vertical expansion by a factor of a. If 0 < a < 1, there is a vertical compression by a factor of a. These rules can be applied to relations as well. What is the effect of replacing the variable y in x + y = 1 with y? There is a vertical expansion by a factor of. The same rule will apply to the x variable. What is the effect of replacing the variable x in x + y = 1 with x 3? There is a horizontal expansion by a factor of 3.
8 Given the graph of x + y = 1 below, sketch the graph of x + y =1. 3 The figure resulting from the two expansions is called an ellipse. List the features of this ellipse. Domain: 3 x 3 Range: y Centre: (0,0) y intercepts: and x intercepts: 3 and 3 In general, the graph of x a + y b contracting) the circle x + y = 1. =1 is an ellipse, which can be obtained by expanding (or General features of the ellipse: Domain: a x a Range: b y b Centre: (0,0) x intercepts: Let y = 0. x a + 0 b = 1 x = ±a y intercepts: Let x = 0. 0 a + y b = 1 y = ±b
9 Note that the ellipse can be elongated in either the vertical or the horizontal direction. (Note: Have students label the diagrams as shown below.) Vertices B 1 and B Vertices A 1 and A The longest chord in the ellipse is called the major axis. The shortest chord in the ellipse is called the minor axis. The major axis and the minor axis intersect at the centre of the ellipse. The endpoints of the major axis are each called a vertex of the ellipse. Two Cases: 1) When a > b, the length of the major axis for x The minor axis will have length b. ) When b > a, the length of the major axis for x The minor axis will have length a a a + y b = 1 is a. + y b = 1 is b. Label the vertices and the major and minor axes on both of the above ellipses.
10. Translating the Graph of an Ellipse Example : Starting with the graph of x 16 + y ( ) 5 = 1, sketch the graph of x 16 ( + y + 3) 5 = 1. Identify the translation that occurs when we: 1) Replace the variable x with (x ). This means a horizontal translation units to the right. ) Replace the variable y with (y + 3). This means a vertical translation 3 units down. Note that x + y 16 5 = 1 can be rewritten as x 4 + y 5 =1. ( x ) 16 ( + y + 3) 5 = 1 Length of major axis: b = 5 = 10 Length of minor axis: a = 4 = 8 Centre: (, 3) Vertices: (, ) and (, 8)
11 3. Standard Form of the Equation of an Ellipse The following is the standard form for an ellipse with centre (h, k). ( x h) a ( + y k) = 1 b There are two cases, depending on whether a > b or b > a. Ellipse with Horizontal Major Axis ( a > b) Ellipse with Vertical Major Axis (b > a) Example 3: Determine the standard form of the conic graphed below. (Exam Specs. #6, p. 6) Centre: ( 3, 5) Length of Major axis: 8 Value of b: 8 = 4 Length of Minor Axis: 4 Value of a: 4 = Equation in Standard Form: (x 3) + (y 5) = 1 or 4 (Sample Exam 00, #6) (x + 3) 4 + (y + 5) 16 = 1
1 Example 4: An ellipse is tangent to the lines x = 5 and x =1. If the centre of the ellipse is on the line y = 3 and the length of the major axis is 1, determine the equation of the ellipse. On the diagram below, sketch the lines x = 5, x =1, and y = 3. Since the ellipse is tangent to the lines x = 5 and x = 1, the length of the horizontal axis must be 5 1 = 6. This means a = 6 and a = 3. Thus, the vertical axis is the major axis of length 1. This means b = 1 and a = 6. By symmetry, the centre of the ellipse must be located midway between the lines x = 5 and x = 1. Therefore, its x coordinate is. Since the centre lies on the line y = 3, its coordinates are (,3). Since the standard form of the equation of an ellipse is ( ) has the equation ( x + ) + y 3 = 1 or 3 6 ( x + ) 9 ( x h) a ( + y 3) 36 ( + y k) = 1, this ellipse b = 1.
13 4. General Form of the Equation of an Ellipse To write the equation ( x + ) 9 and expand the binomial squares. ( + y 3) 36 ( x + ) 9 ( x + ) 36 9 = 1 in general form, we clear the equation of fractions, ( + y 3 ) = 1 36 ( y 3) + 36 = 36 36 4( x + ) + y 3 ( ) = 36 4(x + 4x + 4) + (y 6y + 9) = 36 4x + 16x + 16 + y 6y + 9 = 36 4x + y + 16x 6y 7 = 0 Note that this result resembles closely the equation for a circle. What is the main difference? The coefficients of x and y are different. The general form of the equation of an ellipse has the following form. Ax + Cy + Dx + Ey + F = 0 where A, C, D, E, and F are real constants with A C and A C > 0 (i.e., A and C have the same sign.) What conic do we have if A = C? a circle Note: We are reserving the real constant B for another conic.
14 The Hyperbola Objectives: To investigate the standard form and the general form of the equation of a hyperbola To graph a hyperbola given its equation To write the equation of a hyperbola given its graph 1. Investigation What happens to the graph of the circle x + y = 16 when the plus sign is replaced with a minus sign? In other words, let s investigate the graph of the relation x y =16. a) Domain First, solve the relation x y =16 for the variable y. y = x 16 y = ± x 16 Thus, we must have x 16 0 and x 16. This means the domain is x 4 and x 4. There is a sort of no man s land between x = 4 and x = 4. b) y intercepts Substitute x = 0 into y = ± x 16. Conclusion: There are no y intercepts. This is consistent with the domain restrictions. c) Range Solve the relation for the variable x. x = 16 + y d) x intercepts x = ± 16 + y Substitute y = 0 into x = ± 16 + y. Conclusion: The x intercepts are ±4. Thus, there are no restrictions on the values of y.
15 e) Symmetry of y = ± x -16 Symmetry about the y axis: Both x and x can produce the same value of y. Symmetry about the x axis: Each value of x will produce values of y. f) Calculating some points Use the equation y = ± x 16 to compute the y values in the table below. x 4 4 5 5 6 6 7 7 y 0 0 ±3 ±3 ±4.47 ±4.47 ±5.745 ±5.745 x 8 8 9 9 10 10 y ±6.98 ±6.98 ±8.06 ±8.06 ±9.165 ±9.165 Plot these points on the grid below. The relation x y =16 is called a hyperbola. Notes: 1) A hyperbola has two distinct, unconnected branches. ) The branches are congruent. 3) There is no curve at all on the interval 4 < x < 4.
16 g) Asymptotic Behaviour Use the equation y = ± x 16 to compute the y values in the table below. x 100 1000 10000 100000 1000000 y 99.919968 999.990 9999.999 99999.9999 1000000 What happens to the y values as the x values get very large? As the x values become very large, the values of x become even larger. Thus, the number 16 becomes more and more insignificant when we subtract it from larger and larger values of x. This means that as x gets larger and larger, the value of x or x itself. x 16 gets closer and closer to So in quadrant I, the branch y = line y = x is an asymptote to the curve y = ± x 16. x 16 gets closer and closer to the line y = x. Thus, the We can get the curve y = touch its asymptote. x 16 as close as we like to the line y = x, but it will never By symmetry, the same behaviour occurs in the other three quadrants. The other asymptote to the curve y = ± x 16 is the line y = x. Sketch the asymptotes y = x and y = x on the grid.. Standard Form for the Equation of a Hyperbola The hyperbola x y =16 can be rewritten in the following form. x 16 y x =1 or y =1 16 4 4 This equation is now in the standard form for the hyperbola. In general, the following is the standard form for the equation of a hyperbola centred at the origin.
17 x a y b = 1 What is the main difference between this equation, and the standard form for the ellipse? the subtraction sign A hyperbola like x Example 1: 4 y 4 =1 where a = b is called by some a rectangular hyperbola. Consider the hyperbola 16x 9y = 144. a) Convert the equation to standard form. Divide both sides of the equation by 144. 16x 144 9y 144 = 144 144 x 9 y 16 = 1 b) Find the range of the relation. Solve the equation for x: x 9 y 16 = 1 x y = 1+ 9 16 x = 9 1+ y 16 x = ±3 1+ y 16 This means there is no restriction on the values of y. The range includes all real numbers. c) Calculate the x intercepts. Substitute y = 0 into
18 x = ±3 1+ y 16 x = ±3 1+ 0 16 = ±3 d) Find the domain of the relation. Solve the equation for y. x 9 y 16 = 1 y 16 = x 9 1 y = 16 x 9 1 y = ±4 x 9 1 Thus the domain restriction is x 9 1 0 x 9 e) Determine the asymptotes of the relation. x 3 or x 3. We will consider what happens to the value of y as the value of x increases without bound. y = ±4 x 9 1 As x gets larger and larger, the number 1 becomes more and more insignificant when we subtract it from larger and larger values of x 9. In other words, as x increases, the value of y = ±4 x 9 1 gets closer and closer to ±4 x 9 or ± 4x 3. Thus, the lines y = ± 4 x x are asymptotes to the curve y = ±4 3 9 1.
19 By symmetry, the same behaviour occurs as the value of x gets more and more negative. f) Graph the hyperbola x 3 y 4 =1. Using the facts that the x intercepts are ±3, and that the asymptotes to the curve are y = ± 4 3 x, we can graph the relation quite quickly. The first step is to sketch the rectangle formed by the lines x = 3, x = 3, y = 4, and y = 4. The second step is to draw and extend the diagonals of the rectangle. This rectangle is important for two reasons. 1) The graph lies entirely within two of the sectors marked out by the rectangle s diagonals. ) The diagonals of the rectangle are the asymptotes to the curve. Since the hyperbola gets closer and closer to the diagonals, you can use these lines as guides in sketching the curves. The third step is to plot the x intercepts and another few points for accuracy. x 4 7 10 4 7 10 y ± 3.53 ± 8.43 ± 1.7 ± 3.53 ± 8.43 ± 1.7 Label the points B 1 (0,4), B (0, 4), A 1 (3,0) and A ( 3,0) The points A 1 (3,0) and A ( 3,0) are each called a vertex. The segment A 1 A joining the vertices is the transverse axis. The segment B 1 B is called the conjugate axis.
0 Example : Graph the relation y 3 x 4 =1 How is this relation similar to the one in Example 1? Since the x and y variables have been interchanged, the relations are inverses. Recall from Chapter 1 how the graphs of y 3 x 4 =1 and x 3 y 4 =1 are related. They are reflections of each other in the line y = x. This hyperbola will thus have its branches extending in the y direction. Step 1 Sketch the rectangle formed by the lines y = 3, y = 3, x = 4, and x = 4. Step Extend the diagonals (asymptotes) of the rectangle. Step 3 Plot the y intercepts. What is the length of the conjugate axis? 6 What is the length of the transverse axis? 8 Identify the vertices. (0,3) and (0, 3) (Remember: the transverse axis joins the vertices of the hyperbola.)
1 3. Translating the Hyperbola As with the circle and ellipse, the hyperbola can be translated horizontally and vertically. Example 3: (y +1) (x ) Graph the hyperbola =1. 3 4 Identify its vertices and its centre. Find the equations of the asymptotes and the lengths of the axes. This hyperbola will be congruent to the hyperbola y x =1. But what is its centre? 3 4 Translate the centre (0,0) units right and 1 unit down to get the new centre (, 1). What is the length of the conjugate axis? 4 = 8 What is the length of the transverse axis? 3 = 6 Our strategy will be to first construct a 6 X 8 rectangle centred at (, 1). Then, draw the diagonals (asymptotes). Finally, draw the axes, plot the vertices, and using the asymptotes as guides, sketch the hyperbola. Vertices: (,) and (, 4) Slopes of the Asymptotes: Notice from the completed diagram that the slopes are 3 4 and 3 4. Equations of the Asymptotes: ( ) = 3 4 1) y 1 (x ) y = 3 4 x + 1 ) ( y 1) = 3 (x ) 4 y = 3 4 x 5
4. Summary of the Standard Form of the Equation of a Hyperbola The hyperbola can be oriented in two directions. a) Hyperbola with Horizontal Transverse Axis (x h) (y k) =1 a b Length of Transverse Axis: A 1 A = a Length of Conjugate Axis: B 1 B = b Centre: (h, k) Slopes of Asymptotes: b a and b a b) Hyperbola with Vertical Transverse Axis: (y k) (x h) =1 a b Length of Transverse Axis: A 1 A = a Length of Conjugate Axis: B 1 B = b Centre: (h, k) Slopes of Asymptotes: a b and a b
3 Example 4: Determine the standard form equation of the conic graphed below. Since the hyperbola is oriented vertically, its standard equation will be of the following form. (y k) (x h) = 1 a b Centre of the hyperbola: ( 3, ) Length of Transverse Axis: 8 Value of a: Slopes of Asymptotes: Value of b: a = 8 a = 4 4 = and 4 = Since the slopes of the asymptotes are a b and a b we have b =. Equation of the conic: (y ) 4 (x 3) = 1 or (y ) 16 (x + 3) 4 = 1 Note: The above answer may also be written in the form (x + 3) 4 (y ) 16 = 1
4 5. General Form of the Equation of a Hyperbola The equation (y +1) (x ) =1 can be expanded and cleared of fractions. 3 4 (y + 1) (x ) = 1 3 4 (y + 1) (x ) 144 144 = 144 9 16 16(y + 1) 9(x ) = 144 16(y + y + 1) 9(x 4x + 4) = 144 16y + 3y + 16 9x + 36x 36 = 144 9x + 16y + 36x + 3y 164 = 0 The general form of the equation of a hyperbola has the following form. Ax + Cy + Dx + Ey + F = 0 where A, C, D, E, and F are real constants with A C < 0 (i.e., A and C have different signs.) Remember: We are reserving the real constant B for another conic. Example 5: Change the following equation to standard form. 4x y +16x + y 1= 0 Completing the square, we get 4(x + 4x + 4) (y y + 1) = 1 + 16 1 4(x + ) (y 1) = 16 (x + ) 4 (y 1) 16 = 1 Example 5: Given the conic 5x +16y +100x 3y 84 = 0, a) Change the equation to standard form; and b) Graph the conic on the grid below.
5 a) b)
6 The Parabola Objectives: To investigate the standard form and the general form of the equation of a parabola To graph a parabola given its equation To write the equation of a parabola given its graph Review Recall from Chapter 1 the transformations on the simple parabola y = x. Graph each function on the grid provided, and indicate the translation(s) you used. y = x y = 1 x y + = (x 3) Vertical compression by a factor of 0.5 3 units right, units down y 1 = 1 (x + ) The Axis of Symmetry: x = Vertex: (,1) Vertical compression by a factor of 0. 5, then translate units left, 1 unit up Standard Form for the Equation of a Parabola With our knowledge of transformations, we can state the standard form for the equation of a parabola.
7 Vertical Axis of Symmetry y k = a(x h) [ or y = a(x h) + k ] a > 0 a < 0 Equation of axis of symmetry: x = h Vertex: (h,k) Horizontal Axis of Symmetry x h = a(y k) [ or x = a(y k) + h ] a > 0 a < 0 Equation of axis of symmetry: y = k Vertex: (h,k)
8 General Form for the Equation of a Parabola Consider the parabola y = (x 3) 5. Its standard form equation can be expanded and simplified. This equation is now in general form. y = (x 3) 5 y = (x 6x + 9) 5 y = x 1x + 18 5 y x + 1x 13 = 0 The following is the general form for the equation of a parabola. Ax + Cy + Dx + Ey + F = 0 where A = 0 and D 0 or C = 0 and E 0 What happens if A = 0 and D = 0? We would lose the x variable entirely. What happens if C = 0 and E = 0? We would lose the y variable entirely. What is the axis of symmetry s orientation if C = 0? vertical This will be a parabola of the family y = x or y = x. A > 0
9 A < 0 What is the axis of symmetry s orientation if A = 0? horizontal This will be a parabola of the family x = y or x = y. C > 0 C < 0 Example 1: Use a graphing calculator to sketch the curve y +10y + 4 x +13 = 0. To graph this relation on the calculator, we must solve for y. To solve for y, we need to complete the square. y + 10y + 4x + 13 = 0 (y + 10y + 5) + 4x + 13 = 5 ( y + 5) = 4x + 1 (y + 5) = 4(x 3) y + 5 = ± 4(x 3) y = ± 4(x 3) 5
30 Sketch the calculator result. Vertex: ( 3, 5) Axis of symmetry: y = 5 Standard form of the equation: (x 3) = 1 4 ( y + 5) Example : Give conditions for the constants A, C, and D such that the following equation is a parabola with a horizontal axis of symmetry. Also, assume that the parabola opens to the right. Ax + Cy + Dx + y = 0 This is a parabola of the x = y family. Thus, we must have the following. A = 0 C > 0 D 0 What happens if both A = 0 and D = 0? Cy + y = 0 y(cy + 1) = 0 Two lines: y = 0 and y = 1 C We call this result a degenerate parabola.